AW involves converting all cash flows into an A value

The AW for one life cycle is the same AW for 2,3,4, or

or an infinite number of life cycles,as long as all cash

flows remain the same in succeeding life cycles (or

change by only the inflation of deflation rate)

An asset has a first cost of $20,000 , an annual operatingcost of $8000 and a $5000 salvage value after 3 years.

Calculate the AW for one and two life cycles at i=10%

AWone= -20,000(A/P,10%,3) –8000 +5000(A/F,10%,3)

= -$14,532

AWtwo=-20,000(A/P,10%,6) –8000 –15,000(P/F,10%,3)(A/P,10%,6)

+5000(A/F,10%,6)= -$14,532

Thus, the AW for one life cycle is the same for all life cycles!!

Not necessary to use LCM for different life alternatives

Example: A company is considering two machines for a certain operation. Machine X

will cost $30,000 with annual costs of $18,000 and will have a $7,000 salvage value

after 4 years. Machine Y will cost $50,000 with annual costs of $16,000 and a $9,000

salvage value after its 6 year life. Which machine should the company purchase if

it uses an interest rate of 12% per year?

Solution:

AWX = -30,000(A/P,12%,4) –18,000 +7,000(A/F,12%,4)

AWY = -50,000(A/P,12%,6) –16,000 + 9,000(A/F,12%,6)

= -$26,412

= -$27,052

The company should purchase machine X

Basic equation is A = Pi

Example: Compare the alternatives below using AW and i=10% per year

First cost, $ 50,000 250,000

C D

Annual cost, $/yr 20,000 9,000Salvage value, $ 5,000 75,000 Life, yrs 5 ∞

Solution: AWC = -50,000(A/P,10%,5) – 20,000 + 5,000(A/F,10%,5)= -$32,371

AWD= -250,000(0.10) – 9,000= -$34,000

Select altern1ative C

At an interest rate of 18% per year, the annual worth of an asset which has a first cost of $50,000, an annual operating cost of $30,000, and a $10,000 salvage value after a 4-year life is closest to:

AW = -50,000 (A/P, 18%, 4) – 30,000 + 10,000 (A/F, 18%, 4)                 = -50,000 (0.37174) – 30,000 + 10,000 (0.19174)                 = -$46,670

Annual Worth Problems

What is the annual worth of an asset that has a first cost of $20,000, an annual operating cost of $1,000, and a salvage cost of $10,000 after 3 years at 10% interest?

Answer: AW = -20000(A/P, 10%,3) – 1000 - 10000(A/F, 10%, 3) = -20000(.40211) - 1000 - 10000(.30211) = -12063.30